Simplify the following expression and state the condition under which the simplification is valid. You can assume that $r \neq 0$. $n = \dfrac{7r}{6(3r + 2)} \times \dfrac{18r + 12}{4r} $
When multiplying fractions, we multiply the numerators and the denominators. $n = \dfrac{ 7r \times (18r + 12) } { 6(3r + 2) \times 4r } $ $ n = \dfrac {7r \times 6(3r + 2)} {4r \times 6(3r + 2)} $ $ n = \dfrac{42r(3r + 2)}{24r(3r + 2)} $ We can cancel the $3r + 2$ so long as $3r + 2 \neq 0$ Therefore $r \neq -\dfrac{2}{3}$ $n = \dfrac{42r \cancel{(3r + 2})}{24r \cancel{(3r + 2)}} = \dfrac{42r}{24r} = \dfrac{7}{4} $